# Why is the left inverse of a matrix equal to the right inverse? [duplicate]

Given a square matrix $A$ that has full row rank we know that the matrix is invertible. So there is a matrix $B$ such that

$$AB=1$$

writing this in component notation,

$$A_{ij}B_{jk}=\delta_{ik}$$

Now, we tend to write $A^{-1}$ instead of $B$ but let's leave it like that for now.

My question is how can we show that $BA=1$? We mechanically jump to the conclusion that if the inverse exists, $AA^{-1}=A^{-1}A=1$ but how to show that? Equivalently why is the left inverse equal to the right inverse? It seems intuitively obvious!

Thanks a bunch, I appreciate.

## marked as duplicate by rschwieb, Davide Giraudo, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 13 '15 at 13:33

• This is not a stupid question at all. It is not that trivial (although not super hard either) to show that if $AB=1$ then $BA=1$. – mickep Jan 13 '15 at 11:55
• – Rahul Jan 13 '15 at 11:58
• It is easier to prove it in the abstract, like in group theory. Also, for the conclusion, you need the condition that the matrix be invertible. Some non-invertible (for instance, rectangular) matrives can have right inverse, but no left inverse (or the oposite). – kjetil b halvorsen Jan 13 '15 at 12:04
• In general, this is not true, and $BA$ is not defined. You need to have that $A,B$ are square matrices. – Ofir Schnabel Jan 13 '15 at 12:09
• @ kjetil b halvorsen: I assumed that $A$ has full row rank in line 1 – Georgy Jan 13 '15 at 12:26

So you have a right inverse, and you know there is a left inverse too, let's say $C$. Then you have: $$1 = AB,$$ and multiplying both sides for C in the left, you get $$C = C(AB) = (CA)B = B,$$ that is, if both exist, they must be equal.

Let the endomorphism

$$\Phi:\mathcal M_n(\Bbb R)\to\mathcal M_n(\Bbb R),\; X\mapsto XA$$ then $\Phi(X)=0\iff XA=0\iff XAB=X=0$ hence $\Phi$ is injective and by the rank-nullity theorem it's bijective hence surjective so there's $B'$ such that $\Phi(B')=I_n\iff B'A=I_n$ which means that $A$ has a left inverse. It's routine to prove that $B=B'$.

Given $AB=I$ then $B=BI=B(AB)=(BA)B$ Since $B=(BA)B$ then $BA=I$.

• You don't know $B$ is invertible. – Pedro Tamaroff Jan 13 '15 at 12:19
• In fact you need $B$ to have a right inverse. – Georgy Jan 13 '15 at 12:30

I just had a thought, using the commutator matrix $C=AB-BA=1-BA$. Post-multiply by $B$ we get

$$CB=B-BAB=B-B=0$$

using $AB=1$. Now $B$ has full row rank and therefore $C=0$ implying that $A$ and $B$ commute. Is that a way to go or did I miss something?

EDIT:

Better yet, just consider the matrix $S=ABA$. $$S=(AB)A=A$$ on the other hand $$S=A(BA)$$ Therefore $$A=A(BA) \Rightarrow A(BA-1)=0 \Rightarrow BA=1$$

Lemma: Suppose there is a $v_1\ne0$ so that $Av_1=0$, then the column space of $A$ has dimension at most $n-1$.

Proof: Suppose there is a $v_1\ne0$ so that $Av_1=0$. Create a basis $\{v_k\}_{k=1}^n$ for $\mathbb{R}^n$ including $v_1$. We can write all vectors in $\mathbb{R}^n$ as $$\sum_{k=1}^nc_kv_k$$ which means we can write all vectors in the column space of $A$ as $$\sum_{k=2}^nc_kAv_k$$ Therefore, the column space of $A$ has dimension at most $n-1$. $$\square$$

Since $AB=I$, the column space of $A$ has dimension $n$. Therefore, $Ax=0\implies x=0$.

If $AB=I$, then $ABA=A$. Therefore, $A(BA-I)=0$. This means that each column of $BA-I$ is $0$. That is, $$BA=I$$

Note that this depend on the vector space having finite dimension. Consider the two linear operators on sequences, $A$ and $B$, where $A$ shifts left: $$(Av)_k=v_{k+1}$$ and $B$ shifts right, filling in with $0$: $$(Bv)_k=\left\{\begin{array}{} 0&\text{if }k=0\\ v_{k-1}&\text{if }k\ge1 \end{array}\right.$$ We have $AB=I$, but $BA\ne I$ since $BA$ sets the first element of any sequence to $0$.

• Depending on the level of the reader, I thought a very basic approach might be useful. – robjohn Jan 13 '15 at 14:36